The vectors AB = 3i - 2j + 2k and BC = i - 2k are the adjacent sides of a parallelogram. The angle between its diagonals is

• $\frac{\mathrm{\pi }}{2}$

• $\frac{\mathrm{\pi }}{3} \mathrm{or} \frac{2\mathrm{\pi }}{3}$

• $\frac{3\mathrm{\pi }}{4} \mathrm{or} \frac{\mathrm{\pi }}{4}$

• None of these

The points whose position vectors are 2i + 3j + 4k, 3i + 4j + 2k and 4i + 2j + 3k are the vertices of

• an isosceles triangle

• Right angled triangle

• Equilateral triangle

• Right angled isosceles triangle

P, Q, R and S are four pots with the position vectors 3i - 4j + 5k, - 4i + 5j + k and - 3i + 4j + 3k respectively. Then, the line PQ meets the line RS at the point

• 3i + 4j + 3k

• - 3i + 4j + 3k

• - i + 4j + k

• i + j + k

$\mathrm{If} \mathrm{a} \ne 0, \mathrm{b} \ne 0, \mathrm{c} \ne 0, \mathrm{a} \times \mathrm{b} = 0 \mathrm{and} \mathrm{b} \times \mathrm{c} = 0, \mathrm{then} \mathrm{a} \times \mathrm{c} = ?$

• b

• a

• 0

• i + j + k

805.

The shortest distance between r = 3i + 5j + 7k + λ(i + 2j + k) and r = - i - j - k + μ(7i - 6j + k) is

• $\frac{16}{5\sqrt{5}}$

• $\frac{26}{5\sqrt{5}}$

• $\frac{36}{5\sqrt{5}}$

• $\frac{46}{5\sqrt{5}}$

A unit vector coplanar with i + j + 3k and i + 3j + k and perpendicular to i + j + k is

• $\frac{1}{\sqrt{2}}\left(\mathrm{j} + \mathrm{k}\right)$

• $\frac{1}{\sqrt{3}}\left(\mathrm{i} - \mathrm{j} + \mathrm{k}\right)$

• $\frac{1}{\sqrt{2}}\left(\mathrm{j} - \mathrm{k}\right)$

• $\frac{1}{\sqrt{3}}\left(\mathrm{i} + \mathrm{j} - \mathrm{k}\right)$

If a and b are two non-zero perpendicular vectors, then a vector y satisfying equations a . y = c (where, c is scalar) and a x y = b is

• ${\left|\mathrm{a}\right|}^2\left[\mathrm{ca} - \left(\mathrm{a} \times \mathrm{b}\right)\right]$

• ${\left|\mathrm{a}\right|}^2\left[\mathrm{ca} + \left(\mathrm{a} \times \mathrm{b}\right)\right]$

• $\frac{1}{{\left|\mathrm{a}\right|}^2}\left[\mathrm{ca} - \left(\mathrm{a} \times \mathrm{b}\right)\right]$

• $\frac{1}{{\left|\mathrm{a}\right|}^2}\left[\mathrm{ca} + \left(\mathrm{a} \times \mathrm{b}\right)\right]$

Three non-zero non-collinear vectors $\hat{\mathrm{a}}, \hat{\mathrm{b}} \mathrm{and} \hat{\mathrm{c}}$ are such that $\hat{\mathrm{a}} + 3\hat{\mathrm{b}}$ is collinear with $\hat{\mathrm{c}}$, while $\hat{\mathrm{c}}$ is $3\hat{\mathrm{b}} + 2\hat{\mathrm{c}}$ collinear with a. Then $\hat{\mathrm{a}} + 3\hat{\mathrm{b}} + 2\hat{\mathrm{c}}$ equals

• 0

• $2\hat{\mathrm{a}}$

• $3\hat{\mathrm{b}}$

• $4\hat{\mathrm{c}}$

If $\hat{\mathrm{a}}, \hat{\mathrm{b}} \mathrm{and} \hat{\mathrm{c}}$ are non-coplanar vectors and if $\hat{\mathrm{d}}$ is such that $\hat{\mathrm{d}} = \frac{1}{\mathrm{x}}\left(\hat{\mathrm{a}} + \hat{\mathrm{b}} + \hat{\mathrm{c}}\right)$ and $\hat{\mathrm{d}} = \frac{1}{\mathrm{y}}\left(\hat{\mathrm{b}} + \hat{\mathrm{c}} + \hat{\mathrm{d}}\right)$ where x and y are non-zero real numbers, then $\frac{1}{\mathrm{xy}}\left(\hat{\mathrm{a}} + \hat{\mathrm{b}} + \hat{\mathrm{c}} + \hat{\mathrm{d}}\right)$ equals to

• 3c

• - a

• 0

• 2a

If a, b and c are vectors with magnitudes 2, 3 and 4 respectively, then the best upper bound of ${\left|\hat{\mathrm{a}} - \hat{\mathrm{b}}\right|}^2 + {\left|\hat{\mathrm{b}} - \hat{\mathrm{c}}\right|}^2 + {\left|\hat{\mathrm{c}} - \hat{\mathrm{a}}\right|}^2$ among the given values is

• 93

• 97

• 87

• 90